The 18th offline real-time how to write reference problem in Python

Click here for the problem http://nabetani.sakura.ne.jp/hena/ord18mafovafo/ Click here for execution results http://ideone.com/ZSJEH2 Click here for examples of answers from other people http://qiita.com/Nabetani/items/373105e7fafd12f5e9fd Click here for the event announcement of the "18th Offline Real-Time Writing" Yokohama Henachoko Programming Study Group http://atnd.org/events/47025 (held on Saturday, February 01, 2014)

#!/usr/bin/env python

# FOLD = { operation:(front side fold, back side fold), ... }
# '0':front side paper, '1':back side paper, 'm':mountain fold, 'V':valley fold
FOLD = {'L':('1V0','1m0'), 'J':('0V1','0m1'), 'Z':('0m1V0','1m0V1'), 'U':('1V0V1','0m1m0'), 'S':('0V1m0','1V0m1'), '':('','')}

solve = lambda operation, side=0: ''.join(solve(operation[1:], int(f)) if f in '01' else f for f in FOLD[operation[:1]][side])

def test(data, correct):
    answer = solve(data)
    print 'xo'[answer==correct], data, correct, answer

0, test( "JZ", "mVVmV" );
1, test( "J", "V" );
2, test( "L", "V" );
3, test( "Z", "mV" );
4, test( "U", "VV" );
5, test( "S", "Vm" );
6, test( "JL", "VVm" );
7, test( "JS", "VmVVm" );
8, test( "JU", "VVVmm" );
9, test( "LU", "mmVVV" );
10, test( "SL", "VVmmV" );
11, test( "SS", "VmVVmmVm" );
12, test( "SU", "VVVmmmVV" );
13, test( "SZ", "mVVmVmmV" );
14, test( "UL", "mVVVm" );
15, test( "UU", "mmVVVVmm" );
16, test( "UZ", "mVVmVVmV" );
17, test( "ZJ", "VmmVV" );
18, test( "ZS", "VmmVmVVm" );
19, test( "ZZ", "mVmmVVmV" );
20, test( "JJJ", "VVmVVmm" );
21, test( "JJZ", "mVVmVVmVmmV" );
22, test( "JSJ", "VVmmVVmVVmm" );
23, test( "JSS", "VmVVmmVmVVmVVmmVm" );
24, test( "JUS", "VmVVmVVmVVmmVmmVm" );
25, test( "JUU", "mmVVVVmmVVVmmmmVV" );
26, test( "JUZ", "mVVmVVmVVmVmmVmmV" );
27, test( "LJJ", "VmmVVVm" );
28, test( "LLS", "VmmVmVVmVVm" );
29, test( "LLU", "mmmVVVmmVVV" );
30, test( "LLZ", "mVmmVVmVVmV" );
31, test( "LSU", "mmVVVmmmVVVVmmmVV" );
32, test( "LSZ", "mVVmVmmVVmVVmVmmV" );
33, test( "LZL", "mmVVmVVmmVV" );
34, test( "LZS", "VmmVmVVmVVmmVmVVm" );
35, test( "LZU", "mmmVVVmmVVVmmmVVV" );
36, test( "SJL", "VVmVVmmmVVm" );
37, test( "SLU", "mmVVVVmmmVVmmmVVV" );
38, test( "SLZ", "mVVmVVmVmmVmmVVmV" );
39, test( "SSU", "VVVmmmVVVmmVVVmmmmVVVmmmVV" );
40, test( "SUJ", "mVVVmVVmmmVmmVVVm" );
41, test( "SUS", "VmVVmVVmVVmmVmmVmmVmVVmVVm" );
42, test( "SZZ", "mVmmVVmVVmVmmVVmVmmVmmVVmV" );
43, test( "UJJ", "VmmVVVmVVmm" );
44, test( "ULU", "mmmVVVmmVVVVmmmVV" );
45, test( "ULZ", "mVmmVVmVVmVVmVmmV" );
46, test( "UUU", "VVmmmmVVVmmVVVVmmVVVmmmmVV" );
47, test( "ZJU", "VVVmmmVVmmmVVVVmm" );
48, test( "ZLS", "VmVVmmVmmVmVVmVVm" );
49, test( "ZSJ", "VVmmVmmVVmmVVVmmV" );
50, test( "ZUJ", "mVVVmmVmmmVVmVVVm" );
51, test( "JJLJ", "mVVVmmVVmVVmmmV" );
52, test( "JLJJ", "VmmVVVmVVmmmVVm" );
53, test( "JLJL", "VmmVVVmVVmmmVVm" );
54, test( "LJJL", "VVmmVmmVVVmVVmm" );
55, test( "LLJJ", "VmmmVVmVVmmVVVm" );
56, test( "SZUS", "VmVVmVVmmVmmVmmVmVVmVVmVVmVVmmVmmVmmVmVVmVVmVVmmVmmVmmVmVVmVVmmVmmVmmVmVVmVVmVVm" );
57, test( "ULLS", "VmmVmmVmVVmVVmmVmVVmVVmVVmmVmmVmVVm" );
58, test( "JJJJZJ", "VmmVVVmmVVmVVmmVVmmmVVmVVmmVVVmmVVmmVmmVVmmmVVmVVmmVVVmmVVmVVmmVVmmmVVmmVmmVVVmmVVmmVmmVVmmmVVm" );
59, test( "JULLLJ", "mmVmmVVmmmVVmVVVmmVmmVVVmmVVmVVVmmVmmVVmmmVVmVVVmmVmmVVVmmVVmVVmmmVmmVVmmmVVmVVmmmVmmVVVmmVVmVV" );
60, test( "LJJJUL", "mVVVmVVmmmVVmVVVmmVmmmVmmVVVmVVmmmVmmVVVmmVmmmVVmVVVmVVmmmVVmVVVmmVmmmVVmVVVmVVmmmVmmVVVmmVmmmV" );
61, test( "LJSJJL", "VVmVVmmVVVmmVmmmVVmVVmmmVVmmVmmVVVmVVmmmVVmmVmmVVVmVVmmVVVmmVmmmVVmVVmmVVVmmVmmVVVmVVmmmVVmmVmm" );
62, test( "LZLLLJ", "mmVmmVVmmmVVmVVmmmVmmVVVmmVVmVVVmmVmmVVmmmVVmVVVmmVmmVVVmmVVmVVmmmVmmVVmmmVVmVVVmmVmmVVVmmVVmVV" );
63, test( "SJJJJL", "VVmVVmmVVVmmVmmVVVmVVmmmVVmmVmmVVVmVVmmVVVmmVmmmVVmVVmmmVVmmVmmmVVmVVmmVVVmmVmmVVVmVVmmmVVmmVmm" );
64, test( "ZLJLJL", "VmmVVVmmVmmmVVmVVmmVVVmVVmmmVVmmVmmVVVmmVmmmVVmmVmmVVVmVVmmmVVmVVmmVVVmmVmmmVVmVVmmVVVmVVmmmVVm" );

Commentary

The front side of the paper is '0', the back side is '1', and the mountain fold is'm'. Let the valley fold be'V'. The L-fold result on the front side is (back, valley, front), so '1V0'. The L-fold result on the back side is (back, mountain, front), so '1m0'. Convert from L-fold to folding result to dictionary data FOLD = {'L': ('1V0', '1m0')}. The result of L folding on the front side can be retrieved as '1V0' with FOLD ['L'] [0]. The result of L folding on the back side is FOLD ['L'] [1], and '1m0' can be taken out. If there is only one L fold, the answer is to extract only the'V'of the crease data of '1V0'. In the case of LL folding, 1 and 0 representing the paper part of '1V0' can be replaced with the result of L folding, respectively. LL fold result = Further L fold operation for L fold result on the front side → Fold FOLD ['L'] [0] further L → Further L-fold operation of '1V0' (L-fold operation for 1 and 0) → FOLD['L'][1]+'V'+FOLD['L'][0] → '1m0'+'V'+'1V0' → '1m0V1V0' Then, collect only the crease characters and answer'mVV'. When this is made into a program, it becomes as follows.

1. Extract the first operation character from the operation string
2. Convert to operation characters Look up the dictionary and convert to the folding result
3. If there is a subsequent operation, perform the second and subsequent operations recursively for the folding results 0 and 1.
4. Finally, the work of removing 0s and 1s will be replaced with'' as an empty operation at the end. This will result in an error when there are no operation characters when the operation character is retrieved with operation [0], but it will be an empty string when there is no operation by writing operation [: 1], and the folding result for the empty character If you register as'': ('','') in the conversion dictionary, '0' and '1' will be replaced with empty strings when there are no more operation characters.

solve = lambda operation, side=0: ...
    #solve function definition. Same as the function using def shown below
    # def solve(operation, side=0):
    #     return ...    
operation[:1]
    #Extract the first operation character in the operation string
    #If operation is an empty string''become
FOLD[operation[:1]][side]
    #Extract the folding string for the first operation string of the operation string
    #If operation is an empty string''become
for f in FOLD[operation[:1]][side]
    #For each character in the fold string
operation[1:]
    #Second and subsequent operation strings
solve(operation[1:], int(f)) if f in '01' else f
    # '0'Or'1'Then recursively perform the second and subsequent operations,'m'Or'V'Then that character
    #Remained in the last operation'0'When'1'Solve('', side)Call''become
''.join(...)
    #Combine the folded character strings of the partial parts without any gaps to make a character string